An oil funnel made of a tin sheet consists of a $10 \, cm$ long cylindrical portion attached to a frustum of a cone. If the total height is $22 \, cm$,the diameter of the cylindrical portion is $8 \, cm$,and the diameter of the top of the funnel is $18 \, cm$,find the area of the tin sheet required to make the funnel.

(N/A) Radius $(r_1)$ of the upper circular end of the frustum part $= \frac{18}{2} = 9 \, cm$.
Radius $(r_2)$ of the lower circular end of the frustum part $=$ Radius of the circular end of the cylindrical part $= \frac{8}{2} = 4 \, cm$.
Height $(h_1)$ of the frustum part $= 22 - 10 = 12 \, cm$.
Height $(h_2)$ of the cylindrical part $= 10 \, cm$.
Slant height $(l)$ of the frustum part $= \sqrt{(r_1 - r_2)^2 + h_1^2} = \sqrt{(9 - 4)^2 + (12)^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \, cm$.
Area of the tin sheet required $=$ Curved Surface Area $(CSA)$ of the frustum part $+$ $CSA$ of the cylindrical part.
Area $= \pi(r_1 + r_2)l + 2\pi r_2 h_2$.
Area $= \frac{22}{7} \times (9 + 4) \times 13 + 2 \times \frac{22}{7} \times 4 \times 10$.
Area $= \frac{22}{7} \times 13 \times 13 + \frac{22}{7} \times 80$.
Area $= \frac{22}{7} \times (169 + 80) = \frac{22}{7} \times 249 = \frac{5478}{7} \, cm^2$.
Area $= 782 \frac{4}{7} \, cm^2$.